What is the integral of 1/(x^3+6x^2+11x+6) ?

The integral of 1/(x^3+6x^2+11x+6) is 1/2 ln|x+1|-ln|x+2|+1/2 ln|x+3|+C

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integral of 1/(x^3+6x^2+11x+6)

\int \frac{1}{x ^{3} + 6 x ^{2} + 11 x + 6} d x

\frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{1}{x ^{3} + 6 x ^{2} + 11 x + 6} d x

Take the partial fraction of

\frac{1}{x ^{3} + 6 x ^{2} + 11 x + 6} : \frac{1}{2 ( x + 1 )} - \frac{1}{x + 2} + \frac{1}{2 ( x + 3 )}

= \int \frac{1}{2 ( x + 1 )} - \frac{1}{x + 2} + \frac{1}{2 ( x + 3 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 ( x + 1 )} d x - \int \frac{1}{x + 2} d x + \int \frac{1}{2 ( x + 3 )} d x

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

\int \frac{1}{2 ( x + 3 )} d x = \frac{1}{2} ln ∣ x + 3 ∣

= \frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣ + C

\int \frac{t ^{2} - 1}{t} d t

\int (8 x^{3} + 3 3 x^{2}) d x

\int \frac{e ^{x}}{( e ^{x} + 1 ) ^{14}} d x

\frac{1}{x ^{2} - 1}

\int \frac{ln ( 1 + x ^{2} )}{x ^{3}} d x

What is the integral of 1/(x^3+6x^2+11x+6) ?
The integral of 1/(x^3+6x^2+11x+6) is 1/2 ln|x+1|-ln|x+2|+1/2 ln|x+3|+C